Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 9;
const int LOG = 20;
const int A = 11;
struct Data {
  int id[A];
  Data() { memset(id, 0, sizeof id); }
};
inline Data merge(const Data &a, const Data &b) {
  Data ret;
  ret.id[0] = min(a.id[0] + b.id[0], 10);
  int tmp[A * 2] = {0};
  for (int i = 1; i <= a.id[0]; ++i) tmp[i] = a.id[i];
  for (int i = 1; i <= b.id[0]; ++i) tmp[i + a.id[0]] = b.id[i];
  sort(tmp + 1, tmp + a.id[0] + b.id[0] + 1);
  for (int i = 1; i <= ret.id[0]; ++i) ret.id[i] = tmp[i];
  return ret;
}
int n, m, q;
struct Ugraph {
  int sz;
  int head[N];
  int to[N << 1];
  int ne[N << 1];
  Ugraph() {
    sz = 1;
    memset(head, 0, sizeof head);
  }
  inline void add_edge(int u, int v) {
    to[sz] = v, ne[sz] = head[u], head[u] = sz++;
    to[sz] = u, ne[sz] = head[v], head[v] = sz++;
  }
} G;
bool vis[N];
int fa[N][LOG];
Data g[N][LOG];
int dep[N];
inline void bfs() {
  queue<int> q;
  q.push(1), vis[1] = true;
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    for (int i = G.head[now]; i; i = G.ne[i])
      if (!vis[G.to[i]]) {
        vis[G.to[i]] = true;
        fa[G.to[i]][0] = now;
        dep[G.to[i]] = dep[now] + 1;
        q.push(G.to[i]);
      }
  }
}
inline void pre() {
  for (int j = 1; j <= 17; ++j)
    for (int i = 1; i <= n; ++i) {
      fa[i][j] = fa[fa[i][j - 1]][j - 1];
      g[i][j] = merge(g[i][j - 1], g[fa[i][j - 1]][j - 1]);
    }
}
inline Data up(int &x, int d) {
  int i = 0;
  Data ret;
  while (d) {
    if (d & 1) {
      ret = merge(ret, g[x][i]);
      x = fa[x][i];
    }
    d >>= 1;
    i++;
  }
  return ret;
}
inline Data solve(int u, int v) {
  Data ret;
  if (dep[u] > dep[v])
    ret = up(u, dep[u] - dep[v]);
  else if (dep[u] < dep[v])
    ret = up(v, dep[v] - dep[u]);
  while (u != v) {
    for (int i = 17; i >= 0; --i)
      if (fa[u][i] != fa[v][i]) {
        ret = merge(ret, g[u][i]);
        ret = merge(ret, g[v][i]);
        u = fa[u][i], v = fa[v][i];
      }
    if (fa[u][0] == fa[v][0]) {
      ret = merge(ret, g[u][0]);
      ret = merge(ret, g[v][0]);
      u = fa[u][0], v = fa[v][0];
    }
  }
  ret = merge(ret, g[u][0]);
  return ret;
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    G.add_edge(u, v);
  }
  bfs();
  for (int i = 1; i <= m; ++i) {
    int c;
    scanf("%d", &c);
    if (g[c][0].id[0] < 10) g[c][0].id[++g[c][0].id[0]] = i;
  }
  pre();
  while (q--) {
    int u, v, a;
    scanf("%d%d%d", &u, &v, &a);
    Data ans = solve(u, v);
    int k = min(ans.id[0], a);
    printf("%d", k);
    for (int i = 1; i <= k; ++i) printf(" %d", ans.id[i]);
    puts("");
  }
  return 0;
}